Solving the code lock riddle with Julia

By: perfectionatic

Re-posted from: http://perfectionatic.org/?p=494

I came across a neat math puzzle involving counting the number of unique combinations in a hypothetical lock where digit order does not count. Before you continue, please watch at least the first minute of following video:

The rest of the video describes two related approaches for carrying out the counting. Often when I run into complex counting problems, I like to do a sanity check using brute force computation to make sure I have not missed anything. Julia is fantastic choice for doing such computation. It has C like speed, and with an expressiveness that rivals many other high level languages.

Without further ado, here is the Julia code I used to verify my solution the problem.

  1. function unique_combs(n=4)
  2.     pat_lookup=Dict{String,Bool}()
  3.     for i=0:10^n-1
  4.         d=digits(i,10,n) # The digits on an integer in an array with padding
  5.         ds=d |> sort |> join # putting the digits in a string after sorting
  6.         get(pat_lookup,ds,false) || (pat_lookup[ds]=true)
  7.     end
  8.     println("The number of unique digits is $(length(pat_lookup))")
  9. end

In line 2 we create a dictionary that we will be using to check if the number fits a previously seen pattern. The loop starting in line 3, examines all possible ordered combinations. The digits function in line 4 takes any integer and generate an array of its constituent digits. We generate the unique digit string in line 5 using pipes, by first sorting the integer array of digits and then combining them in a string. In line 6 we check if the pattern of digits was seen before and make use of quick short short-circuit evaluation to avoid an if-then statement.